Question

If the two ends of the major axis of an ellipse are \((5,0)\) and \((-5,0)\) and one focus lies on the line \(3x-5y-9=0\), then its equation is:

• A. \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\)
• B. \(\dfrac{x^2}{16}+\dfrac{y^2}{25}=1\)
• C. \(\dfrac{x^2}{25}+\dfrac{y^2}{34}=1\)
• D. \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\)

Hint:

For ellipses: \[ c^2=a^2-b^2 \] Always identify whether the major axis is horizontal or vertical before writing the equation.

Answer 1

The Correct Option is A

Understanding the Concept: The standard equation of an ellipse centered at the origin with major axis along the x-axis is: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] where: \[ \text{Vertices}=(\pm a,0) \] and the foci are: \[ (\pm c,0) \] with: \[ c^2=a^2-b^2 \]
Step 1: Finding the value of \(a\). The ends of the major axis are: \[ (5,0),\ (-5,0) \] Therefore: \[ a=5 \] Hence: \[ a^2=25 \]
Step 2: Finding the focus coordinate. Since the ellipse is horizontal, the focus lies on the x-axis. So focus is: \[ (c,0) \] Given that this point lies on: \[ 3x-5y-9=0 \] Substituting \(y=0\): \[ 3x-9=0 \] \[ 3x=9 \] \[ x=3 \] Thus: \[ c=3 \]
Step 3: Finding \(b^2\). Using: \[ c^2=a^2-b^2 \] Substituting values: \[ 3^2=5^2-b^2 \] \[ 9=25-b^2 \] \[ b^2=16 \]
Step 4: Writing the equation of ellipse. Substituting \(a^2=25\) and \(b^2=16\): \[ \frac{x^2}{25}+\frac{y^2}{16}=1 \]